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1

Answer:

I predict that the product will be 1-chloro-2-nitrocyclohexane.

Explanation:

We would expect #"HCl"# to add to 1-nitrocyclohexene according to Markovnikov's rule and form 1-chloro-1-nitrocyclohexane:

Mark

However, this mechanism puts a positive charge on the carbon atom next to another positive atom:

Mech

In this case, the addition is more likely to be anti-Markovnikov.

anti-Mark

This avoids two adjacent + charges in the intermediate and forms a secondary carbocation.

I predict the product will be 1-chloro-2-nitrocyclohexane.

3

Answer:

The compound is 3-methyl-2-butanone.

Explanation:

To begin with the IR-abosorption, #1720# #cm^-1# peak shows that this compound has a carbonyl group.
The molecule of #C_5H_10O# has a double bonding in the carbonyl group, and has no #C=C# double bonding.

Then, let's proceed to the NMR spectrum.
[1] A peak near #1.10# ppm corresponds to #-CH_3# group. This is a doublet, so there will be a single proton next to it.
[2] A peak at #2.10# ppm is for the #-(C=O)-CH_3#. There is no proton in its neighbor.
[3] The fact that the peak at #2.50# ppm is a septet tells us there are six protons in the adjacent point. This might be for the #-(C=O)-Ccolor(red)H-(CH_3)_2#.

Therefore, the structure will be
#CH_3-(CO)-CH-(CH_3)_2#. The IUPAC name is 3-methyl-2-butanone.
https://pubchem.ncbi.nlm.nih.gov/compound/3-methyl-2-butanone#section=Top

Let's check the answer:
https://www.chemicalbook.com/SpectrumEN_563-80-4_1HNMR.htm

Here is a chemical shift table.(PDF)
https://staff.aub.edu.lb/~tg02/nmrchart.pdf#search=%27proton+NMR+table%27

5

Answer:

Warning! Long Answer. The compound is propionic anhydride.

Explanation:

Preliminary analysis

You know the formula is #"C"_6"H"_10"O"_3"#.

An alkane with six carbon atoms has the formula #"C"_6"H"_14"#.

The degree of unsaturation #U# is

#U = (14-10)/2 = 4/2 = 2#

Therefore, the compound contains two rings and/or double bonds.

#""^1"H NMR"#

The spectrum has 10 protons and only two peaks. The molecule must have a symmetrical structure.

A peak with 2 neighbours and aone with 3 neighbors corresponds to an ethyl group (#"C"_2"H"_5#, triplet-quartet pattern).

The #"6H:4H"# pattern tells us there are two ethyl groups.

However, I think you have the assignments reversed. The methyl group should have the smaller chemical shift.

andromeda.rutgers.edu

A methyl group is normally at 0.9 ppm. Something is pulling it downfield
to 1.2 ppm.

A methylene group is normally at 1.3 ppm. Something is pulling it downfield
to 2.5 ppm.

We see from the table that a #"CH"_3# next to a #"C=O"# is shifted downfield to 2.2 ppm (a shift of 1.3 ppm).

We could expect a similar 1.3 ppm shift for a #"CH"_2# group; from 1.2 ppm to 2.5 ppm.

This is just where the #"CH"_2# group appears.

We now know that the partial structure is

#"CH"_3"CH"_2"(C=O)"# + #"(O=C)CH"_2"CH"_3#

These fragments add up to #"C"_6"H"_10"O"_2"#.

There is only one #"O"# atom left to insert. It must go in the middle:

#"CH"_3"CH"_2"(C=O)-O-(O=C)CH"_2"CH"_3#

The compound is propionic anhydride.

Confirmation:

1. The compound has two double bonds.

2. Three #""^13"C"# NMR signals tell us there are three different carbon environments,

image.slidesharecdn.com

We should expect to see

  • #"CH"_3color(white)(mm)# at 10 ppm
  • #"CH"_2 color(white)(mm)# at 30 ppm
  • #"(C=O)O"# at 170 ppm

We see peaks at 8 ppm, 28 ppm, and 170 ppm. This is consistent with propionic anhydride.

3. The #""^13"C"# NMR spectrum of propionaldehyde is

www.chemicalbook.com

3

Answer:

Warning! Long Answer.
1. 4 #"C"_3# axes, 6 #σ_text(d)# mirror planes, and 3 #"S"_4# improper axes
2. 1 #"C"_text(3v)# axis and 2 #σ_text(v )# planes

Explanation:

Point group for #"CF"_4#

#"CF"_4# has a tetrahedral geometry. It belongs to point group #"T"_text(d)#, like methane.

CF4

It has four #"C"_3# axes, three #"C"_2# axes, three #"S"_4# improper axes, and six #σ_text(d)# mirror planes.

(a) #"C"_3 # axes

If you place the model on a table and look "down" the vertical #"C-F"# bond, we can see that each bond is a #"C"_text(3)# axis of rotation.

You can rotate the molecule 120°# about the axis and get an indistinguishable configuration of the molecule.

C3

(b) #"C"_2# axes

The image below shows one of the four #"C"_text(2)# axes.

C2

The axis bisects a #"C-F"# bond angle.

You can rotate the molecule 180° about this axis and get an indistinguishable configuration of the molecule.

(c) #"σ"_text(d)# planes

The image below shows the two of the six #σ_text(v)# planes.

Sigma

One is the plane of the screen. The other is a plane perpendicular to the screen and bisecting an #"F-C-F"# bond angle.

(d) #S"_4# axes.

An #"S"_text(n)# improper rotation is a rotation of #(360°)/n# about a #C"_n# axis, followed by reflection in a plane perpendicular to that axis.

S4
(Adapted from slideshare.net)

There are four of these axes in the molecule.

#bb(1.)color(white)(m)"CF"_4# to #"CClF"_3#

The structure of #"CClF"_3# is

CClF3

We see that the molecule has a #"C"_text(3v)# axis and two #σ_text(v )# planes of symmetry.

#"C"_text(3v)# axes

The three-fold axis is more obvious if you look down the #"C-Cl"# with the #"Cl"# atom closest to your eye.

C3v

You can rotate the molecule 120° left or right and get a new configuration that is indistinguishable from the original.

#sigma_text(v)# planes

3σv

Each #σ_text(v)# includes a #"Cl-C-F"# bond. Here the mirror plane is the plane of the screen.

A reflection converts the molecule into an indistinguishable form of the original.

In going from #"CCl"_4# to #"CClF"_3#, the molecule has lost four #"C"_3# axes, six #σ_text(d)# mirror planes, and three #"S"_4# improper axes

**#bb(2.)color(white)(m)"CClF"_3# to #"CBrClF"_2#

The structure of the molecule is

CBrClF2

Thus, it is in point group #"C"_text(s)#. It has only a #σ_text(v )# plane of symmetry.

#σ_h# plane

σh

The plane contains the #"Br-C-F"# bonds and bisects the #"Cl-C-F"# bond angle#.

One half of the molecule is a mirror image of the other half.

Thus, in going from #"CClF"_3# to #"CBrClF"_2#, the molecule has lost a #"C"_text(3v)# axis and two #σ_text(v )# planes of symmetry.

2

Answer:

Some examples are: #CO_2#, #O_2#, #CH_3# DNA, non-polar amino acids

Explanation:

Covalent bonds are common in the molecules of living organisms. This bonds are created is by sharing electrons. The more electrons they share the stronger the bond will be.
Non-polar covalent bonds appear between two atoms of the same element or between different elements that equally share electrons.

  • #O_2# and #CO_2#
    #O_2# is non-polar because the electrons are equally shared between the two oxygen atoms. The same case is with #CO_2#.
    https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/110/2016/05/02211833/Figure_02_01_11jpg
    Breathe in and out! When you inhale, #O_2# comes in your lungs, and when you exhale, #CO_2# goes out. This process happens because #CO_2# and #O_2# are nonpolar molecules.
    www.bioweb.lu

  • #CH_3#
    Our intestinal gas/ flatus is produced as a byproduct of bacterial fermentation in our colon.
    doctordaliah.files.wordpress.com
    Click here for: Why is #CH_4# a non-polar molecule?
    https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/110/2016/05/02211833/Figure_02_01_11jpg

  • DNA
    Each strand of the DNA is polar, but each strand with its polarity neutralizes the molecule of the DNA in total.
    images.slideplayer.com

  • NON-POLAR amino acids #-># their R-groups, white on the picture below, hydrocarbon alkyl groups (alkane branches) or aromatic rings, exception: benzene ring in the amino acid Tyrosine which is polar. Aminoacids are monomers that form proteins.
    Alanine, Cysteine, Glycine, Isoleucine, Leucine, Methionine, Phenylalanine, Proline, Tryptophan, Valine

www.hammiverse.com

2

Answer:

More or less the same as Aldoses..

Explanation:

The biochemical class of compounds known as Sugars can roughly be divided in two categories: Aldoses and Ketoses.

They typically contain between 3 and 7 Carbon atoms:

  • 3:#rarr# Trioses
  • 4:#rarr# Tetroses
  • 5:#rarr# Pentoses
  • 6:#rarr# Hexoses
  • 7:#rarr# Heptoses

7 seems to be the maximum: there are no (known) sugars containing 8 or more Carbon atoms, as they are too unstable.

If we start with the simplest ones we find there are 3 Trioses:

  • Dihydroxyacetone;
  • D-Glyceraldehyde;
  • L-Glyceraldehyde.

Have a look at the glyceraldehydes :

enter image source here

D-Glyceraldehyde and L-Glyceraldehyde are identical, except that they are mirror-images of each other, a bit like a pair of gloves: one left-handed, the other right-handed. The glyceraldehydes are the starting point for all Aldoses, so we'll forget about them for the rest.

Next, the difference between D-Glyceraldehyde and Dihydroxyacetone :

BillytheKid Studios

Next, we number the Carbon atoms from top to bottom:
enter image source here

enter image source here
If we go from Triose to Hexose, we find that at each step an extra H-C-OH-group has crept in:

enter image source here

Observe that in each of the ketoses shown, the one-but-last H-C-OH - group "points" in the same direction: This is the group that determines D- or L-configuration. All the ketoses shown are D-configs. Starting with #C_3#, the other (asymmetrical) carbon centres determine the Identity of the ketose, e.g whether it is Xylulose or Ribulose.

NOTE: The last but one C-centre doesn't change this identity, only the D/L-configuration!!!!!!!!!!!!!!!

Footnote : I didn't include the Heptoses: theoretically there could be 8 of them, but there aren't that many around, and they don't seem to be very important in biochemical respect anyway.....

4

ISOMER 1: n-bromobutane

The primary (#1^@#) bromobutane (#"C"_4"H"_9"Br"#) has the electrophilic carbon marked below.

#"H"_3"C"-"CH"_2-"CH"_2-stackrel(delta^+)(stackrel("*")"C")"H"_2-stackrel(delta^-)"Br"#

You said that you are using #"OH"^(-)(aq)#, which would do a nucleophilic backside-attack on the primary carbon marked #"*"#.

Due to the low steric hindrance around that carbon (no non-H substituents are on #"C"^"*"#), it is susceptible to a second-order substitution (#"S"_N2#).

Therefore, the nucleophile and substrate will both participate in a reaction that has no intermediate, giving a rate law of:

#r(t) = k["OH"^(-)]["C"_4"H"_9"Br"]#

The reason why an intermediate doesn't form is that primary carbocations are very unstable.

The electron density from a #"C"-"H"# bond would spread out via hyperconjugation to stabilize the positive charge on the central carbon as shown below by sending electron density into an empty #2p_z# orbital:

http://wps.prenhall.com/

The more alkyl groups you have surrounding the #"C"^"*"#, the more stabilized it is because alkyl groups are electron-releasing (electron-donating, giving a so-called "inductive effect").

Instead of an intermediate, a transition state forms, which is simply going to look halfway between the reactant and the product. It will be a trigonal bipyramidal geometry around the central carbon!

#" "" "" "" "" "" "" "" "color(white)(.)"Br"#
#" "" "" "" "" "" "" "" "color(white)(.)vdots#
#"H"_3"C"-"CH"_2-"CH"_2-stackrel("*")"C"^(delta^+)"H"_2#
#" "" "" "" "" "" "" "" "color(white)(.)vdots#
#" "" "" "" "" "" "" "color(white)(....):ddot"O":^(delta^(-))#
#" "" "" "" "" "" "" "" "" ""|"#
#" "" "" "" "" "" "" "color(white)(.....)"H"#

Of course, you should draw it with the proper bond angles. Be sure to include the partial charges.

ISOMER 2: tert-butyl bromide

You can tell then that tert-butyl bromide (#("H"_3"C")_3"CBr"#) is more substituted.

#" "" "" ""CH"_3#
#" "" "" ""|"#
#"H"_3"C"-"C"-"Br"#
#" "" "" ""|"#
#" "" "" ""CH"_3#

It has a tertiary (#3^@#) central carbon, so it has more steric hindrance, i.e. it has more things blocking a nucleophile from coming in for a backside-attack, so only one molecule can participate in the reaction (the substrate).

Hence, a first-order substitution (#"S"_N1#) will occur, and a planar carbocation intermediate will form, giving a rate law of:

#r(t) = k_1[("CH"_3)_3"CBr"]#

(where the first step is slow)

while the intermediate looks like this:

#" "" "" ""CH"_3#
#" "" "" ""|"#
#"H"_3"C"-"C"^(+)#
#" "" "" ""|"#
#" "" "" ""CH"_3#

(Of course, you should be drawing this with the proper bond angles!)

This is stabilized by the three #"CH"_3# electron-releasing alkyl groups around the central carbon, spreading electron density out to stabilize the positive charge via hyperconjugation as before.

This stabilizing effect is much stronger than in a primary carbocation.

[Being planar, the intermediate will lead to a racemic mixture of #R//S# stereoisomers if the alkyl group(s) around the central carbon are all different.]

1

Longest carbon chain is to be selected. If there exist options then longest carbon chain containing unsaturation is to be selected in preference to the saturated chain as shown in example 2,
4,4-dipropylhept-1-en-6-yne.
enter image source here

2

Answer:

In this compound there are 4 optical isomers; this is calculated by multiplying the number of chiral centres by two.

Explanation:

The number of optical isomers in a compound is determined by the number of chiral centres in it. A chiral centre is a carbon atom that is bonded to four different molecules or atoms. Each chiral centre will result in two different optical isomers.

So, to work out the number of chiral centres, draw out the displayed formula of the compound, and circle/highlight any carbons in the compound that have four different molecules attached to them (to check this, try drawing the molecule as a tetrahedral shape around your chosen chiral centre; each bond from the central C atom should go to a unique compound). Then, just multiply the number of chiral centres by two to give the number of optical isomers.

Diagrams:

So below is your molecule (I have used MolView to sketch this):
http://molview.org/?cid=139799
To work out which carbons are chiral, you just have to check each one separately. A chiral carbon is usually not on a branched group, or on the end of a chain, so often you can eliminate those pretty quickly. Below is the molecule with the chiral centres highlighted in yellow:

http://molview.org/?cid=139799
Theory of optical isomerism:

Stereoisomers are compounds that have the same structural formulae, but due to the lack of rotation around a double carbon bond (aka #pi#-bond) will form slightly different compounds where a structure is above the molecular plane in one isomer, and below in the other.

Optical isomers (aka enantiomers) follow a similar principle, except instead of thinking about differences in location above and below a double bond, we are thinking about the orientation around a tetrahedral carbon centre.

The rule is that if each of the four substituents that are bonded to the central carbon are different, then it is possible for a stereoisomer of this compound to exist which is a mirror image of this compound. It is a form of stereoisomerism because both compounds have the same structural formulae, but cannot be superimposed on top of each other no matter how you rotate them.

The easiest way to get a feel for this is to draw a tetrahedral molecule, with (any) four different bonding groups; then imagine a mirror line next to the compound and draw its mirror image on the other side (this effect can be achieved by using a real mirror). Then, try and superimpose one molecule on top of the other - it should not be possible.