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Answer:

Any molecule that has a nonzero vector sum of dipole moments is said to be polar and have a dipole moment

Explanation:

A dipole moment refers to slight opposite charges on opposite sides of a bond. The resulting bond is said to be polar; it has a positive pole and a negative pole, much like a bar magnet.

In order to determine if a particular bond is polar or not, one must look for the electronegativity of each atom. Pauling's electronegativity is a measure of how strong a particular atom pulls electrons towards it in a bond. The value of the difference between their electronegativities (#DeltaEN#) determines how polar a bond is.

If:
#0<= DeltaEN <=0.4#, the bond is nonpolar.

#0.4<##DeltaEN##<=1.8#, the bond is polar

#DeltaEN>1.8#, the bond is ionic

Consider the bonds in #H_2O#,
https://socratic.org/questions/how-many-dipoles-are-there-in-a-water-molecule

There is one oxygen bonded to two hydrogens in one water molecule. Based on the difference in electronegativites for the bonds, it is clearly a polar molecule
#EN_O=3.44#
#EN_H=2.20#
#DeltaEN=3.44-2.20=1.24#

In the figure above, the #delta# symbol indicates an area of partial charge on the atom. Note that they are not full charges as in ions, but partial charges due to a difference in electron density at each "pole". The arrow in the figure indicates the direction of electron density and the slight negative charge #delta^-#and the cross indicates an area of electron deficiency and the slight positive charge #delta^+#. This difference in charges is called a dipole moment and it is a vector quantity; it has magnitude and direction.

Notice that the water molecule has an overall dipole moment that points straight up towards the oxygen. This is because a dipole moment of a molecule depends on the vector sum of the bond dipoles.

Consider #CO_2#,
http://butane.chem.uiuc.edu/pshapley/GenChem2/B3/1.html

As you can see, the #DeltaEN# for the #C-O# bond is within the polar range. However, since #CO_2# is a linear molecule, the dipoles point in opposite directions and the vector sum of the two is equal to zero. #CO_2# is nonpolar.

Answer:

Here's what I get.

Explanation:

When there is more than one permissible IUPAC name, the first name is preferred.

Ethers

There are two systems of IUPAC names for ethers.

1. Substitutive names.

The ethers are named as alkoxyalkanes, with the senior component selected as the parent compound.

Thus, #"CH"_3"OCH"_2"CH"_2"CH"_3# is 1-methoxypropane.

2. Functional group names.

The ethers are named as alkyl alkyl ethers, with the alkyl groups in alphabetical order followed by the class name ether, each as a separate word.

Thus, #"CH"_3"OCH"_2"CH"_2"CH"_3# is methyl propyl ether.

Esters

Esters are named as alkyl alkanoates.

The name of the alkyl group is written first, followed by the name of the acid with the ending -ic acid replaced by the ending -ate.

Thus, #"CH"_3"CH"_2"COO-CH"_3# is methyl propanoate (two words).

Cyanides (nitriles)

There are three systems of IUPAC names for nitriles.

1. As alkanenitriles.

The ending -nitrile is added to the name of the alkane with the same number of carbon atoms.

Thus, #"CH"_3"CH"_2"C≡N"# is propanenitrile.

2. If the compound is considered to be formed from a carboxylic acid with a "trivial name" (#"RCOOH → RC≡N"#), the ending -ic acid is changed to -onitrile.

Thus, #"CH"_3"CH"_2"C≡N"# is also called propionitrile (from propionic acid).

3. As alkyl cyanides.

The name of the alkyl group precedes the class name cyanide.

Thus, #"CH"_3"CH"_2"-C≡N"# is ethyl cyanide (two words).

It can be considered as basically spatial crowding.


Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).

As an example, consider the reaction seen below, which one might hope is #"S"_N2#:

http://www.masterorganicchemistry.com/

Here, the nucleophile is cyanide (#""^(-):"CN"#) and the electrophile is the central carbon on the alkyl halide (tert-butyl bromide).

We should predict that reaction does not work via an #"S"_N2# mechanism, based on the idea of steric hindrance---the three #"CH"_3# groups are blocking the cyanide from performing its backside-attack.

https://qph.ec.quoracdn.net/

(LEFT: steric hindrance; RIGHT: reduced steric hindrance)

That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).

Hence, this reaction proceeds more easily as a first-order mechanism (e.g. #"S"_N1# or #E1#), since the #"Br"^(-)# has the time (during the slow step) to come off on its own.

The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:

#r(t) ~~ k_1["Br"^(-)]#

(Depending on the choice of solvent, either #"S"_N1# or #E1# could occur.)

You'll have to be able to draw these out to name them. It is impossible to name by inspection of their chemical formulas.

  1. I've highlighted the longest hydrocarbon chain, which you would have had to identify. #(a)# is therefore a kind of hexane, while #(b)# is a kind of heptane (hex = six, hept = seven carbons in the main chain).

  2. Now you simply count from each end to determine the lowest possible set of carbon indices, and account for duplicate functional groups using prefixes. Alphabetize them afterwards.

NOTE: the smallest index should be the first point of difference between two sets of numberings.

There are four methyl groups in #(a)#, while there is one methyl and one ethyl group in #(b)#.

Counting from the left:

#(a)# is then denoted #ul"2,2,3,3-tetramethylhexane"#.

#(b)# is then denoted #"3-ethyl-6-methylheptane"#.

Counting from the right:

#(a)# is then denoted #"4,4,5,5-tetramethylhexane"#.

#(b)# is then denoted #ul"5-ethyl-2-methylheptane"#.

The first name for #(a)# is correct since its indices are minimized.

The second name for #(b)# is correct since the first encountered functional group with the lowest index is methyl, but since ethyl comes before it alphabetically, we rearrange these while keeping the indices as they are.

#(b)# is more difficult, so remember that example.

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

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

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  • Al E. answered · 3 days ago
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