(P) (R-I)/(( Excess ))(Q) Aq ·Ag2 O⟶(R) Δ⟶ T+S If (T) is simplest hydrocarbon alkene and P is

step 1 Hello everyone. Let us see the following question. There is a compound P when treated with alkal

(P) (R-I)/(( Excess ))(Q) Aq ·Ag2 O⟶(R) Δ⟶ T+S If (T) is...

$$ \text { (P) } \frac{R-I}{(\text { Excess })}(Q) \stackrel{\mathrm{Aq} \cdot \mathrm{Ag}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{R}) \stackrel{\Delta}{\longrightarrow} \mathrm{T}+\mathrm{S} $$ If (T) is simplest hydrocarbon alkene and $\mathrm{P}$ is $\mathrm{R}^{\prime}-\mathrm{NH}_{2}$, then choose the correct option(s): (A) $\mathrm{P} \frac{\mathrm{NaNO}_{2}}{+\mathrm{HCl}} \stackrel{\mathrm{TsCl}}{\mathrm{Py}} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{T}$ (B) In the above reaction, $\mathrm{R}^{\prime}$ is methyl group (C) P gives base soluble compound on reaction with Hinsberg reagent (D) Increasing basic strength order in aq. phase will be $\left(\mathrm{NH}_{3}<\mathrm{P}<\mathrm{S}\right)$.

$$
\text { (P) } \frac{R-I}{(\text { Excess })}(Q) \stackrel{\mathrm{Aq} \cdot \mathrm{Ag}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{R}) \stackrel{\Delta}{\longrightarrow} \mathrm{T}+\mathrm{S}
$$
If (T) is simplest hydrocarbon alkene and $\mathrm{P}$ is $\mathrm{R}^{\prime}-\mathrm{NH}_{2}$, then choose the correct option(s):
(A) $\mathrm{P} \frac{\mathrm{NaNO}_{2}}{+\mathrm{HCl}} \stackrel{\mathrm{TsCl}}{\mathrm{Py}} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{T}$
(B) In the above reaction, $\mathrm{R}^{\prime}$ is methyl group
(C) P gives base soluble compound on reaction with Hinsberg reagent
(D) Increasing basic strength order in aq. phase will be $\left(\mathrm{NH}_{3}<\mathrm{P}<\mathrm{S}\right)$.

Key Concepts

Basicity Trends in Amines and Their Derivatives

The basic strength of amines depends on the electron availability at the nitrogen atom. In general, ammonia is a strong base, followed by primary amines which are slightly less basic due to the electron-donating effects of alkyl groups. However, when the amine is converted into a sulfonamide (as in reactions involving sulfonyl chlorides), the strong electron-withdrawing effect of the sulfonyl group markedly reduces the electron density on the nitrogen, thus substantially lowering its basicity. This concept is crucial when comparing the basic strength among ammonia, amines, and their sulfonamide derivatives in an aqueous environment.

Diazotization of Primary Amines

Diazotization involves the reaction of primary amines with nitrous acid (typically generated from sodium nitrite and a mineral acid) to form diazonium salts. In the case of aliphatic amines, these diazonium intermediates are usually unstable and decompose via loss of nitrogen, leading to further transformations such as substitution by nucleophiles or elimination reactions that can yield alkenes.

Tosylation of Amines

Tosylation refers to the conversion of an amine into its corresponding tosyl derivative by reaction with tosyl chloride (TsCl). This process temporarily converts the amine into a more stable or selectively reactive intermediate by installing a tosyl (p-toluenesulfonyl) group, which not only reduces the basicity of the nitrogen but also creates a good leaving group, facilitating subsequent elimination or substitution reactions.

Hofmann Elimination Reaction

Hofmann elimination is a type of elimination reaction where a quaternary ammonium salt, typically derived from an amine that has been modified (often via tosylation or alkylation), undergoes dehydrohalogenation in the presence of a strong base, especially under heated and alcoholic conditions. The reaction usually leads to the formation of the less substituted (Hofmann) alkene, highlighting the influence of steric factors on the elimination pathway.

Hinsberg Test for Amines

The Hinsberg test is a diagnostic reaction used to differentiate between primary, secondary, and tertiary amines based on their behavior when reacted with sulfonyl chlorides such as benzenesulfonyl chloride. Primary amines typically form sulfonamides that exhibit specific solubility characteristics in aqueous base, serving as an important qualitative tool to classify the type of amine present.

Transcript

0:00 Hello everyone.

00:01 Let us see the following question.

00:03 There is a compound p when treated with alkali iodide in excess should give a compound q which when treated with acquiesce silver oxide should give r and then this r when heated should give us t and yes.

00:23 Now we know that the structure of p is given as r nh2 and there is a hint.

00:30 Given t is the simplest alkyme so t should be the simplest alkyme so that is nothing but c h2 double bond c h2 so if t is the simplest alken that means we can go back to the reaction sequence so that means we can go backward so t is c h2 double bond c h2 and yes should be having the three units of a remaining c2 h5 connected to nitrogen.

01:02 So, this is happened when heated r.

01:05 So, r should be c2h5 totally four times on nitrogen.

01:11 Okay.

01:13 Now, if it's so, then corresponding p should be c2h5, nh2.

01:19 The r should be here, nothing by ethyl group.

01:24 Now, when treated with excess of ri, that is once again c2h5, i, in excess, we have to get c2h5 four times on nitrogen and iod.

01:39 When treated with aqueous silver oxide, we have to get as n plus and oh minus ammonium hydroxide.

01:52 So here we have done cope elimination.

01:55 In cope elimination, one of the etail group breaks and undergoes elimination to form corresponding ethyne.

02:02 So, let us understand the options one by one.

02:06 Option a says that, so by the way, this is t, this is s, this is r, this is q and this is p.

02:15 Option a says that when p treated with the na in no2, so when p is what now, c2, h5 and nh2, when treated with freshly prepared na in no2, plus hc, cl should give us what c2h5 diazoneum, n2 plus cl minus, which on treated with alcoholic k oea should give t.

02:45 Yes, because n2 is eliminated, nitrogen is eliminated, we get ch2 double bond ch2.

02:53 So this statement is going to be the true statement.

02:57 So let's see the statement b, where in the above reaction, r is a methyl group.

03:03 No, that's the wrong statement...

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