14.
Answer is E
R-MgBr will reduce secondary ketone into OH with an addition of R group onto the C-O carbon
Drawing all the products out will help you see the structure more clearly and understand why all
three reactions give the same product.
15.
Answer is A
(CH3)2CHCH2CH2Br + Mg/ether - (CH3)2CHCH2CH2(MgBr) (MgBr will be a good leaving
group)
(CH3)2CHCH2CH2(MgBr) + CO2 in H3O+ - (CH3)2CHCH2CH2-CO2H (which is a carboxylic acid)
16.
Answer is D
sources of a nucleophile that can attack
the + end of the C=O double bond in aldehydes and ketones. The methyl group
of CH3MgBr will be added into the carbon (with oxygen) also.
CH3MgBr is a Grignard reagent and is a
17.
Answer is E
This one is a Grignard reagent reactions again as above. Reactions B & C can be explained as
above. Reaction A is reaction between Grignard reagent and ester with general understanding
can be shown below. Note that the circled part of ester was eliminated during the reaction and
the 2 carbon groups of Grignard were added instead.
18.
The answer is C (3-pentanol)
Methyl formate is
Grignard reagent is CH3-MgBr
The reaction is Grignard with ester as shown above.
19.
Answer
1. PBr3 (SN2 reaction / substitute 1 bromine in place of OH)
2. Mg/ether to give CH3MgBr (See Question 15)
3. HCHO/H3O+ (CH3MgBr react with HCHO (formaldehyde) see question 16)
20.
1. Grignard: Phenyl-MgBr + Butanal
2. Switch places between Grignard and Aldehyde on the picture to have
Grignard: Propyl-MgBr + benzaldehyde