Doubt from alkane

Ans given is a

What's the problem Acid chloride is reduced to primary alcohol by NaBH4. It's his property.
For your confirmation

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But nabh4 reduces acid chloride to aldehyde only na as it is not strong reducing agent

Lialh4 reduces it @Gaurav_2020_3 @Shwetanshu_2018 bhaiya please confirm

and given is d

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@Gaurav_2020_3 my sir told that nabh4 reduces it only to aldehyde now I am confused please help is something I missing

@Bhuvanitha_2020 @Shwetanshu_2018 @Samarth_2020

@Vibhu_2020 I didn't understand what your query is. Which reaction are you asking..?

Where do you have problem in this?
Amides have been converted to amines and usual ester reduction and Acid chloride to Alcohol

In 17th one my doubt is that I have written in my notes that it gets stop at only aldehyde further it is not reduced . And in ans as u told that is also further gets reduced to alcohols. Now I am confused what I have missed in my notes I have uploaded (my notes pic) please just tell that @Gaurav_2020_3 whether have I wriiten incorrect in my notes

Bro can u solve 13 th one @Samarth_2020

I am sure about it, It's wrong in your notes. You can strongly take a stand about it talk to your subject teacher about it " politely ".
I provided the solution of Q 13 in my previous post. Please tell me whether you feel uncomfortable in that question

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NaBh4 don’t reduce Ester/amides/carboxylic acids.
Otherwise it reduces any other carbonyl compounds to alcohol.
Pic from my notes is below:

For restricted reduction DIBAL-H is used .

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The only diff between LiAlh4 and NaBh4 is the first one reduces Ester/amides/acids.
Otherwise both reduce carbonyl to alcohol.

If anyone find anything wrong pls say me

@Vibhu_2020 say in what part of q you have doubt . (In q no 13)

See bhuvanitha ans given is d and my doubt where that oh has gone I have reduced that ketone to alcohol so my ans coming is a please see it @Bhuvanitha_2020