# Can anyone help me with these cyclization questions?

Oct 2, 2016

Here are my suggestions.

#### Explanation:

(a) Malonate $\text{S"_"N} 2$ Displacement

(b) Malonate Conjugate $\text{S"_"N} 2$ Displacement

(c) Epoxide Ring Opening

Oct 2, 2016

Question 1: probably the easiest, this is standard carbanion and enolate chemistry.

#### Explanation:

The substrate is:

$C l \left(C {H}_{2}\right) C H = C H {\left(C {H}_{2}\right)}_{4} O C \left(= O\right) C {H}_{2} C \left(= O\right) O M e$; the methylene protons between the 2 carbonyl groups to the right are enolizable to give:

$C l \left(C {H}_{2}\right) C H = C H {\left(C {H}_{2}\right)}_{4} O C \left(- {O}^{-}\right) = C H C {O}_{2} M e$.

This is the enolate, and this is the carbanion:

$C l \left(C {H}_{2}\right) C H = C H {\left(C {H}_{2}\right)}_{4} O C \left(= O\right) - {C}^{-} \left(H\right) C {O}_{2} M e$.

This reacts as the carbanion to effect ring closure at the allylic carbon on the other end of the chain to eliminate halide.

The way it is drawn on the paper is much more intuitive than the way I have drawn it in this editor, because you can see the potential $C - C$ bond formed between the enolizable carbon and the allylic position.

The point is that $R O \left(O =\right) C - C {H}_{2} C {O}_{2} M e$ protons are enolizable by methoxide, and once formed this carbanion can react at the other end of the chain to form a $C - C$ with halide eliminated.