Question

Show the product of treating the following \(\gamma\)-lactam with each reagent. CN1CCCC1=O (a) \(\frac{\mathrm{H}_{2} \mathrm{O}, \mathrm{HCl}}{\text { heat }}\) (b) \(\frac{\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}}{\text { heat }}\) (c) \(\frac{1 . \mathrm{LiAlH}_{4}}{2 . \mathrm{H}_{2} \mathrm{O}}\)

Short answer

Answer: The products obtained from treating the given γ-lactam with the following sets of reagents are: (a) C(NH)CH2CH2CH2COOH, (b) C(NH)CH2CH2CH2COOH, and (c) C(NH)CH2CH2CH2CH2OH.

Step-by-step solution

Protonation of the carbonyl

The first step is the protonation of the carbonyl oxygen by \(\mathrm{HCl}\). This makes the carbonyl carbon more susceptible to attack by nucleophiles, such as water molecules, due to the positive charge on the oxygen. The intermediate product would be \(\mathrm{CN1CCCC1(OH_{2})^{+}=O}\).

Ring opening

The protonated γ-lactam is unstable and undergoes a ring-opening reaction, prompted by the attack of the negatively charged nitrogen lone pair on the carbonyl carbon. This results in the structure \(\mathrm{C(NH)CH_{2}CH_{2}CH_{2}COOH}\). (b) \(\frac{\mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}}{\text { heat }}\)

Nucleophilic attack

This reaction involves the nucleophilic attack of hydroxide ion (\(\mathrm{OH^{-}}\)) on the carbonyl carbon. This results in the formation of an alkoxide ion intermediate: \(\mathrm{CN1CCCC1(O^{-})=O}\).

Protonation and ring opening

The alkoxide ion intermediate is unstable and will grab a proton (H\(^{+}\)) from the solvent, followed by ring opening similar to the mechanism in (a). The final product would be the same as in (a): \(\mathrm{C(NH)CH_{2}CH_{2}CH_{2}COOH}\). (c) \(\frac{1 . \mathrm{LiAlH}_{4}}{2 . \mathrm{H}_{2} \mathrm{O}}\)

Reduction of carbonyl

\(\mathrm{LiAlH}_{4}\) is a strong reducing agent, which will reduce the carbonyl group of the γ-lactam to an alcohol, forming an intermediate \(\mathrm{CN1CCCC1(OH)}\).

Ring opening

Despite the intermediate not being charged like in the previous cases, the ring-strain in the γ-lactam structure promotes the ring-opening reaction in a similar manner to the previous examples. This results in the structure \(\mathrm{C(NH)CH_{2}CH_{2}CH_{2}CH_{2}OH}\) as the final product.