Match the following: List I (Reactants) 1. HC ≡CH+H2 O Hg^2 *⟶ 2. PhCH3 (1 . Cl2, Δ)/(2 . H2 O /

Name: Problem 187, Chapter 28: A Complete Resource Book in Chemistry for JEE Main
Uploaded: 2022-01-29T17:21:24-08:00
Duration: 7 min 50 s
Channel: Rajesh Singh
Description: Match the following: List I (Reactants) 1. HC ≡CH+H2 O Hg^2 *⟶ 2. PhCH3 (1 . Cl2, Δ)/(2 . H2 O / OH^-) 3. CH3-CH=CH2+CO+H2 [Co(CO)4]2⟶ 4. H2 C=CH2+O2 PdCl2, CuCl2H2 O List II (Products) (i) CH3 CHO (ii) PhCHO (iii) CH3 CH2 CH2 CHO The correct matching is: The correct matching is: 1 2 3 4 (a) (iii) (ii) (i) (i) (b) (i) (i) (ii) (iii) (c) (ii) (iii) (i) (i) (d) (i) (ii) (iii) (i)

step 1 Hello everyone here is a matching type question. So in the matching type of question we have a l

Match the following: List I (Reactants) 1. HC ≡CH+H2 O Hg^2...

Match the following: List I (Reactants) 1. $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{Hg}^{2 *}}{\longrightarrow}$ 2. $\mathrm{PhCH}_{3} \frac{1 . \mathrm{Cl}_{2}, \Delta}{2 . \mathrm{H}_{2} \mathrm{O} / \mathrm{OH}^{-}}$ 3. $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{CO}+\mathrm{H}_{2} \stackrel{\left[\mathrm{Co}(\mathrm{CO})_{4}\right]_{2}}{\longrightarrow}$ 4. $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{O}_{2} \stackrel{\mathrm{PdCl}_{2}, \mathrm{CuCl}_{2}}{\mathrm{H}_{2} \mathrm{O}}$ List II (Products) (i) $\mathrm{CH}_{3} \mathrm{CHO}$ (ii) $\mathrm{PhCHO}$ (iii) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$ The correct matching is: The correct matching is: 1 2 3 4 (a) (iii) (ii) (i) (i) (b) (i) (i) (ii) (iii) (c) (ii) (iii) (i) (i) (d) (i) (ii) (iii) (i)

Match the following:
List I (Reactants)
1. $\mathrm{HC} \equiv \mathrm{CH}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{Hg}^{2 *}}{\longrightarrow}$
2. $\mathrm{PhCH}_{3} \frac{1 . \mathrm{Cl}_{2}, \Delta}{2 . \mathrm{H}_{2} \mathrm{O} / \mathrm{OH}^{-}}$
3. $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{CO}+\mathrm{H}_{2} \stackrel{\left[\mathrm{Co}(\mathrm{CO})_{4}\right]_{2}}{\longrightarrow}$
4. $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{O}_{2} \stackrel{\mathrm{PdCl}_{2}, \mathrm{CuCl}_{2}}{\mathrm{H}_{2} \mathrm{O}}$
List II (Products)
(i) $\mathrm{CH}_{3} \mathrm{CHO}$
(ii) $\mathrm{PhCHO}$
(iii) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$
The correct matching is:
The correct matching is:
1 2 3 4
(a) (iii) (ii) (i) (i)
(b) (i) (i) (ii) (iii)
(c) (ii) (iii) (i) (i)
(d) (i) (ii) (iii) (i)

Key Concepts

Wacker Oxidation

The Wacker oxidation is a catalytic process that converts alkenes into carbonyl compounds using a palladium(II) catalyst along with a co-catalyst, typically copper(II) chloride, in the presence of water. This reaction is notable for its mild conditions and its ability to selectively oxidize alkenes to aldehydes or ketones, making it a versatile tool in organic synthesis.

Hydroformylation Reaction

Hydroformylation, also known as the oxo process, is the catalytic addition of a formyl group and a hydrogen atom to an alkene. Carried out in the presence of metal carbonyl catalysts (like cobalt carbonyl complexes), this reaction transforms alkenes into aldehydes by incorporating carbon monoxide and hydrogen. The resulting products are valuable intermediates in the synthesis of various chemicals.

Hydration of Alkynes

This concept refers to the addition of water to a carbon–carbon triple bond, typically in the presence of a catalyst such as mercuric ion. The reaction transforms an alkyne into a carbonyl compound (often a ketone or aldehyde) through Markovnikov addition, and it is widely used in organic synthesis to introduce oxygen functionality.

Benzylic Oxidation

Benzylic oxidation involves the selective oxidation of a benzylic position—that is, the carbon atom directly attached to an aromatic ring. Under appropriate conditions, such as using halogenation followed by hydrolysis, a methyl group attached to an aromatic ring can be converted into a formyl group, yielding an aromatic aldehyde. This reaction is a key transformation in modifying or functionalizing aromatic compounds.

Transcript

00:01 Hello everyone here is a matching type question.

00:03 So in the matching type of question we have a list one or a column one which shows some reactants.

00:13 Okay.

00:14 So in the list two we are getting some products.

00:19 List two is giving some products.

00:25 Okay.

00:26 So now let us see the question one.

00:29 In question one there is a ethyne.

00:32 So see.

00:33 Triple bond ch, which is graded with water in the presence of mercurus.

00:41 So this reaction is called kucharose reaction.

00:45 So let us do the mechanism of the reaction.

00:47 So if the bond is breaking, okay, loan pay is here.

00:52 This is positive.

00:53 So the respective intermediate what we get is ch, double bond, positive charge here, c and h and loan pay here.

01:01 So, h plus adds here, h plus adds to here, and which leads to the formation of ch positive charge double bond, ch2.

01:14 Now, the oh nucleophile in the first step attacks the carbocatao.

01:21 So, the product obtained here is ch -o -h double bond ch2.

01:29 We know very well in carbonyl groups having unsaturation totomerism plays a very important role.

01:38 So on totomerism, we know that hydrogen migrate to the alpha place and double bond shift to the carbonyl oxygen.

01:48 So we have to get here c double bond oh, that is aldehyde and this side ch3.

01:54 So, we get here ethanol, ethanol.

01:59 So, this option is however given in 1.

02:03 So, for 1, right choice will be, so in list 1, list 1 itself that is given as ch3, cho.

02:13 So, therefore, for 1, correct match, it should be first one.

02:23 So, let us do the second reaction.

02:25 In second reaction, it is given tall v.

02:32 H3 which has to be chlorinated in the presence of heat.

02:37 This is step one.

02:39 So in the presence of heat, chlorination should happen on side chain because it follows free radical mechanism which eliminates one unit of hl.

02:51 Now in step two, it has to be treated with h2o in the presence of bis.

02:58 So this is going to remove the cl.

03:01 So, here we are going to do substitution nucleophilic.

03:05 So, oh would substitute the same carbon.

03:09 So, therefore, the resulting molecule obtained here is benzile alcohol.

03:16 Now, oxidized medium, that means it gets oxidizes.

03:22 So, the molecule obtained here should be benzarddehyde, c -h -o.

03:27 So, therefore, for question 2, question 2, right choice is ph -c -h -o, that is option 2 itself.

03:44 Now, let us see the third question.

03:47 In third question, it is given ch3, c -h double bond, c -h -2, treated with coo plus h2 in the presence of cobalt tetracarboneal complex...

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