Chemistry · Haloalkanes And Haloarenes · NEET
It gives a carboxylic acid with ONE extra carbon. The carbanion R- attacks the carbon of CO2 (O=C=O). First you get a magnesium carboxylate salt (R-COO- Mg+ X), the intermediate. Then acidic water (H3O+) adds an H to give R-COOH. So CO2 is a trick to add a -COOH group. Example: CH3MgBr + CO2 then H3O+ gives CH3COOH (acetic acid).
It depends on the carbonyl. The R- adds to the C=O carbon, then H3O+ gives an alcohol. Rule: Formaldehyde (HCHO) gives a PRIMARY (1°) alcohol. Any other aldehyde (RCHO) gives a SECONDARY (2°) alcohol. A ketone (like acetone) gives a TERTIARY (3°) alcohol. So acetone + CH3MgCl gives tert-butyl alcohol, a 3° alcohol.
The R-Mg carbon is like a strong base (a hungry carbanion). Water has an O-H bond, and that H is slightly positive. The carbanion grabs this H and becomes R-H (an alkane). So the Grignard is used up giving no useful product. That is why Grignard reactions MUST be done in dry ether with NO water. This is why moisture is the enemy.
Active hydrogen means an H attached to O, N, S, or a terminal alkyne carbon (-OH, -NH2, -NH, -SH, -C≡C-H). These H-atoms are slightly acidic. A Grignard is a very strong base, so it just takes this H and turns into R-H (alkane). No addition happens. NEET loves testing this: if the other molecule has any O-H, N-H or ≡C-H, the answer is usually just the alkane.
Formaldehyde is HCHO (two H on the carbonyl carbon). R- adds there, and after H3O+ you get R-CH2-OH, a primary (1°) alcohol. Any bigger aldehyde R'CHO has one H, so you get R-CH(OH)-R', a secondary (2°) alcohol. Simple ladder to remember: HCHO gives 1°, RCHO gives 2°, ketone gives 3°.
D2O acts exactly like water - it has O-D bonds. The carbanion grabs the D atom, so R-D is formed. This is a common NEET trick to put a deuterium exactly where the metal was. Example: C6H5MgBr + D2O gives C6H5-D (deuterobenzene). So D2O is a smart way to label one specific carbon.
Reaction between acetone and methylmagnesium chloride, followed by hydrolysis, will give:
In the reaction RMgX + CO2 (dry ether) -> Y (H3O+) -> RCOOH, the intermediate Y is:
Identify X: p-Br-C6H4-Cl (1.0 mol) + Mg (1.0 mol) in dry ether -> Intermediate, then D2O -> X
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Dry ether has no O-H bonds and its oxygen lone pairs help stabilise the R-MgX. Any water or moisture would give its H to the carbanion and destroy the reagent, turning it into a useless alkane R-H. So everything must be perfectly dry.
Look at the carbonyl. Formaldehyde (HCHO) gives 1° alcohol. Any other aldehyde (RCHO) gives 2° alcohol. A ketone gives 3° alcohol. Count the carbons already on the C=O carbon and add one from the Grignard.
No. It adds only once across one C=O of O=C=O, giving a carboxylate salt (R-COO- Mg+ X). After H3O+ you get one carboxylic acid RCOOH with just one new -COOH group.
Only the alkane R-H forms. Alcohols (O-H) and amines (N-H) have active hydrogen. The strong basic carbanion grabs that H, so no addition happens. For example, cyclohexyl-MgBr + an amine gives cyclohexane.
Yes, very. NEET repeatedly asks Grignard products - CO2 to acid, ketone to 3° alcohol, and the active-H trap where water/amine/alcohol gives only the alkane. Knowing these three patterns lets you answer most questions in seconds.
