Please help with resulting products of given reactions

Question

what would be the resulting products from the following reactions:

(reagents shown in italics)

a) 2-methylcyclopentanoic acid ---CH₃OH/H₂SO₄---> ?

b) 4-methylbenzoic acid ---SOCl₂---> ?

c) cyclohexanecarbonyl chloride ---CH₃CH₂NH₂---> ?

d) 2-(4-methylcyclohexyl)acetamide ---1. LiAlH₄/2. H₂O---> ?

e) methyl 3-methylbutanoate ---1. DIBAH/2. H₃O⁺---> ?

f) 2-phenylethanol ---acetic anhydride/pyridine---> ?

g) ethyl benzoate ---1. CH₃MgBr/2. H₃O⁺---> ?

h) methyl 2,4-cyclopentadien-1-ylacetate ---1. LiAlH₄/2. H₃O⁺---> ?

Answer 1

(a) Acid catalyzed reaction of CH₃OH with a carboxylic acid to form a methyl ester.

(b) The reagent SOCl₂ converts a carboxylic acid (R–CO₂H) to an acyl chloride ((R–COCl).

(OH is replaced by Cl.)

(c) Amine reacts with acyl chloride to produce an amide (O=C–NH linkage), but ½ of the amine gets converted to CH₃CH₂NH₃⁺ so 2 equivalents of CH₃CH₂NH₂ should be used for the reaction.

(d) The amide (–O=C–NH₂) in the molecule gets reduced to an amine (–CH₂–NH₂) by LiAlH₄.

(e) Might be intended as DIBAH reduction of an ester to an aldehyde, but the first step should be at –78 °C.

Possible (though unlikely) that it's a trick question and the answer is a 1° alcohol (which is what would actually happen if the first step were carried out at a higher temperature.)

(f) The alcohol PhCH₂CH₂OH reacts with (CH₃CO)₂O (acetic anhydride) to produce an acetate ester

PhCH₂CH₂O–COCH₃.

(g) CH₃MgBr adds twice to an ester to give a 3° alcohol.

Reason: the ketone that forms after CH₃MgBr displaces ⁻OCH₂CH₃ is more reactive than the ester it is formed from, so it reacts again. (Two equivalents of CH₃MgBr should be used for all of the ester to react.)