Question

Draw a structural formula for the principal product formed when benzoyl chloride is treated with each reagent. (a) OC1CCCCC1 (b) CCCCO (c) \(\mathrm{SH}\), pyridine (d) \(\mathrm{NH}_{2}\) (two equivalents) (e) CCC(=O)ON (f) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}\), then $\mathrm{H}_{3} \mathrm{O}^{+}$ (g) COc1ccc(N)cc1 (h) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}\) (two equivalents), then \(\mathrm{H}_{3} \mathrm{O}^{+}\)

Short answer

Question: Identify the principal product formed when benzoyl chloride reacts with each of the following reagents: (a) tetrahydrofuran (THF) (b) n-butanol (c) thiol (SH), pyridine (d) two equivalents of ammonia (NH2) (e) N-hydroxysuccinimide (NHS) (f) lithium dimethylcuprate (CH3)2CuLi, then H3O+ (g) methyl 4-aminobenzoate (h) phenylmagnesium bromide (C6H5MgBr) (two equivalents), then H3O+. Answer: (a) C6H5COO(C1CCCCC1) (b) C6H5COOCCCC (c) C6H5COSH (d) C6H5CONH2 (e) C6H5COOCCC(=O)ON (f) C6H5COCH(CH3)2 (g) C6H5CONHCOC6H4 (h) C6H5COPh

Step-by-step solution

Exercise (a)

React benzoyl chloride with OC1CCCCC1. The reagent is tetrahydrofuran (THF), which has the structural formula "OC1CCCCC1". In this reaction, THF behaves as a nucleophile, and benzoyl chloride is the electrophile. The oxygen in THF will attack the carbonyl carbon of benzoyl chloride, while the chloride will leave as a leaving group. The principal product formed is C6H5COO(C1CCCCC1).

Exercise (b)

React benzoyl chloride with CCCCO. The reagent is n-butanol, which has the structural formula "CCCCO". Similar to the reaction in Step 1, n-butanol will act as a nucleophile attacking the carbonyl carbon of benzoyl chloride, and the chloride will leave as a leaving group. The principal product formed is C6H5COOCCCC.

Exercise (c)

React benzoyl chloride with \(\mathrm{SH}\), pyridine. Pyridine is a base and will facilitate the reaction between benzoyl chloride and thiol (SH). Benzoyl chloride reacts with the nucleophilic thiol (SH) in an electrophilic and nucleophilic substitution reaction, where the sulfur attacks the carbonyl carbon, and the chloride leaves as a leaving group. The principal product formed is C6H5COSH.

Exercise (d)

React benzoyl chloride with \(\mathrm{NH}_{2}\) (two equivalents). Ammonia (NH2) will act as a nucleophile and reacts with the carbonyl carbon of benzoyl chloride, with the chloride leaving as a leaving group. Since two equivalents of ammonia are used, the reaction will proceed further and form an amide. The principal product formed is C6H5CONH2.

Exercise (e)

React benzoyl chloride with CCC(=O)ON. The reagent is N-hydroxysuccinimide (NHS), which has the structural formula "CCC(=O)ON". NHS acts as a nucleophile, reacting with the carbonyl carbon of benzoyl chloride, with the chloride as the leaving group. The principal product formed is C6H5COOCCC(=O)ON.

Exercise (f)

React benzoyl chloride with \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}\), then \(\mathrm{H}_{3} \mathrm{O}^{+}\). The reagent is lithium dimethylcuprate, which is a nucleophile that performs an addition reaction with the carbonyl carbon of benzoyl chloride. Subsequently, hydrolysis takes place in the presence of H3O+. The principal product formed is C6H5COCH(CH3)2.

Exercise (g)

React benzoyl chloride with COc1ccc(N)cc1. The reagent is methyl 4-aminobenzoate, which has the structural formula "COc1ccc(N)cc1". The nitrogen in the amino group acts as a nucleophile and reacts with the carbonyl carbon of benzoyl chloride, while the chloride leaves as a leaving group. The principal product formed is C6H5CONHCOC6H4.

Exercise (h)

React benzoyl chloride with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}\) (two equivalents), then \(\mathrm{H}_{3} \mathrm{O}^{+}\). The reagent is phenylmagnesium bromide, which is a nucleophile and will react with the carbonyl carbon of benzoyl chloride in a Grignard reaction. Since two equivalents are used, this reaction will form a diaryl ketone. The presence of H3O+ neutralizes the charges, forming a neutral product. The principal product formed is C6H5COPh.