Would LiAlH4 be a viable reagent for converting methyl propanoate to 1-propanol + methanol?

Other possible reagents i can choose from are:

NaBH4

Fe/HCL

PCC

Sn/HCL

....I would think LiALH4 would be the preferred reagent because i read this on the internet "LiAlH4 is a powerful reducing agent which reduces carboxylic acids to primary alcohols in good yields"

 

I know cold dilute sodium bicarbonate could be used for a chemical test to distinguish between benzoic acid and benzamide. Could cold dilute sodium hydroxide be used also?

 

 

This other question in the textbook has got me stumped too.

When treated with sodium iodide, a solution of (S)-2-iodooctane in acetone.....(more than 1 correct answer)

1. does not react

2. loses its optical activity

3. gives both ® - and (S) -2-iodooctane

4. is converted to ® -2-iodooctane

I'm really struggling with this question and don't know where to start.

 

 

Also, what would be the name for the compound C7H12 that reacts with ozone followed by hydrogen peroxide to give heptane-2,6-dione as the only organic product?

 

Thanks.

Forget that last question, any help on the others would be very much appreciated.

If im not wrong ester can't undergo reduction !!!

 

C7H12 ???? what is that ??

theres no such hydrocarbon !! it cant be a aliphatic nor a aromatic cpd !!

esters can undergo reduction......i found the answer to that question anyway.

any thoughts on the others?

ester undergoes reduction ..

but does not yield carboxylic acid and alcohol !!!

 

It yields 2 alcohols only !!

 

To get carboxylic acid and alcohol from ester ,

u can undergo the reverse reaction of esterification !!!

stop confusing me

i need answers!!

If im not wrong ester can't undergo reduction !!!

You are wrong' date=' esters undero reduction with LiAlH4

 

C7H12 ???? what is that ??

theres no such hydrocarbon !! it cant be a aliphatic nor a aromatic cpd !!

It is heptyne.

 

Would LiAlH4 be a viable reagent for converting methyl propanoate to 1-propanol + methanol?

Other possible reagents i can choose from are:

NaBH4

Fe/HCL

PCC

Sn/HCL

....I would think LiALH4 would be the preferred reagent because i read this on the internet "LiAlH4 is a powerful reducing agent which reduces carboxylic acids to primary alcohols in good yields"

You are correct on this.

 

I know cold dilute sodium bicarbonate could be used for a chemical test to distinguish between benzoic acid and benzamide. Could cold dilute sodium hydroxide be used also?

As long as the sodium hydroxide does not react wtih the amide, then you can use it.

 

As for your (second) last question, I don't have the time to think about it now, I am (stupidly) surfing the forums when I have a biochemistry exam in less than 3 hours

This other question in the textbook has got me stumped too.

When treated with sodium iodide' date=' a solution of (S)-2-iodooctane in acetone.....(more than 1 correct answer)

1. does not react

2. loses its optical activity

3. gives both ® - and (S) -2-iodooctane

4. is converted to ® -2-iodooctane

I'm really struggling with this question and don't know where to start.[/quote']

 

Here's some starting points to think about. How would I^- react with (S)-2-iodooctane? What does this reaction do to the stereochemistry of the product? And finally, would I^- have any preference for reacting with (S)-2-iodooctane or ®-2-iodooctance?