Chemistry

Get insights from 6.9k questions on Chemistry, answered by students, alumni, and experts. You may also ask and answer any question you like about Chemistry

Follow Ask Question
6.9k

Questions

0

Discussions

25

Active Users

0

Followers

New answer posted

9 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

ANSWER:

1. PhMgBr and then H3O+

 

 

New answer posted

9 months ago

0 Follower 53 Views

A
alok kumar singh

Contributor-Level 10

ANSWER:     

1. The 2, 4-dinitrophenylhydrazone of benzaldehyde

 


New answer posted

9 months ago

0 Follower 27 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 
 

 

 

New answer posted

9 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution:

 

 

New answer posted

9 months ago

0 Follower 74 Views

A
alok kumar singh

Contributor-Level 10

Answer:
 
iii.
 

 

 

New answer posted

9 months ago

0 Follower 40 Views

A
alok kumar singh

Contributor-Level 10

(i) Cyanohydrin:

Cyanohydrins are those organic compounds having the formula RR“2C (OH)CN, wherever R and R“2 can be alkyl or aryl groups.

Aldehydes and ketones react with compound (KCN) within the presence of excess cyanide (NaCN) as a catalyst to field organic compounds.

(ii) Acetal:

Acetals are gem - dialkoxy alkanes within which 2 alkoxy teams groups attached to the terminal atom. One bond is connected to an associate degree alkyl, whereas the opposite is connected to the hydrogen atom.

When aldehydes are treated with 2 equivalents of a monohydric alcohol within the presence of dry HCl gas, hemiacetals are produced which are furth

...more

New answer posted

9 months ago

0 Follower 31 Views

A
alok kumar singh

Contributor-Level 10

(i) The +I effect of –CH3 group increases the electron density on the O-H bond. Therefore, the release of proton becomes difficult. On the other hand, the -I effect of F decreases the electron density on the O-H bond. Therefore, the proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.

(ii) F has stronger -I effect than Cl (as fluorine is more electronegative than chlorine). Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is a stronger acid than CH2ClCO2H.

(iii) Inductive effect decreases with increase in distance (the more will be the distance of the inductive group from

...more

New answer posted

9 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

(i) Ethylbenzene

In the presence of nascent oxygen i.e. [O] along with KMnO4/ KOH (it is a very strong oxidizing agent which oxidizes the whole alkyl group to carboxylic salt) followed by hydrolysis leads to the formation of benzoic acid. Carbon dioxide and water are formed as byproducts.

(ii) Acetophenone

The oxidation with alcoholic KMNO4 followed by hydrolysis leads to the formation of benzoic acid. The reaction is given below:

(iii) Bromobenzene

Bromobenzene is first reacted with magnesium in presence of dry ether to give an intermediate which on reaction with solid carbon dioxide followed by hydrolysis leads to the formation of benz

...more

New answer posted

9 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

(i) Ph CH2 COOH The compound contains carboxylic acid as the functional group and Ph means a phenyl group. The longest chain contains 3 carbon atom with phenyl group of carbon 3 so the IUPAC name of the compound is 3-phenylpropanoic

(ii) (CH3 ) 2 C=CHCOOH

The longest chain contains 4 carbon atom with a methyl group as a substituent and a carboxylic acid as a functional group. The no. Of the chain starts from the carbon atom of the –COOH group with double bond on carbon 2 so the IUPAC name of the compound is 3-Methylbut-2-enoic acid.

(iii) The longest chain has 5 carbon atom (all are saturated) which are cyclic with the carboxylic gr

...more

New answer posted

9 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

(i) Cyclopentanone reacts with hydroxylamine to give cyclopentanone oxime. As you can see the cyclopentanone has ketone functional group with oxygen attached to carbonyl carbon and the hydroxylamine has two H atoms attached to N. And hence in the product, this O and two H atoms are removed and form a water molecule.

Remember if any aldehyde or ketone bearing compound reacts with primary or secondary amine then for finding the product formed simply remove the oxygen atom from the aldehyde or ketone functional group and two H atoms attached to N in amines and add a double bond between C and N.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.