How can the but- 1 -ene to but- 2 -ene conversion be effected ?

But-1-ene on reacting with hydrochloric acid, the major product formed is 2-chlorobutane according to Markonikoff's rule. 2-chlorobutane on reacting with Alc . KOH , dehydrohalogenation takes place and But-2-ene is formed as the major product. CH 3 − CH 2 − CH = CH 2 → HCl CH 3 − CH 2 − CH ( Cl ) − CH 3 CH 3 − CH 2 − CH ( Cl ) − CH 3 → Alc . KOH CH 3 − CH = CH − CH 3

Name and state the rule that governs the formation of major product, when 2 -Bromopentane is heated with alcoholic solution of potassium hydroxide?

2 -Bromopentane, when heated with alcoholic solution of potassium hydroxide, it undergoes dehydrohalogenation to give Pent- 2 -ene as a major product. This is according to Saytzeff rule, which can be stated as, in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Thus, 2 -Bromopentane gives Pent- 2 -ene as the major product.

Which is the major product obtained when 2 -Bromopentane is heated with alcoholic solution of potassium hydroxide?

2 -Bromopentane, when heated with alcoholic solution of potassium hydroxide, it undergoes dehydrohalogenation to give Pent- 2 -ene as a major product. This is according to Saytzeff rule, which can be stated as, in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Thus, 2 -Bromopentane gives Pent- 2 -ene as the major product.

2 -Bromobutane is optically active. Explain the stereo-chemical aspect of S N 1 reaction of 2 -Bromobutane with OH - ions.

2 - bromobutane is a chiral molecule as it contains an asymmetric carbon atom therefore, it is optically active In case of optically active alkyl halides, SN 1 reactions are to go with racemisation. In this reaction the carbo cation formed in the slow step is sp 2 hybridised and is planar (achiral). The attack of the nucleophile take place from either side of the plane of carbocation resulting in a mixture of products, one having the same configuration (the – OH attaching on the same position as halide ion) and the other having opposite configuration (the – OH attaching on the side opposite to halide ion). Example: Hydrolysis of optically active 2-bromobutane, which results in the formation of ( ± ) - butan - 2 - ol.

Aryl halides are less reactive towards nucleophilic substitution reactions. Write any two reasons for the less reactivity of aryl halides.

Aryl halides are less reactive in nucleophilic substitution reactions. The two reasons for less reactivity are: Resonance effect: The lone pair of electrons of halogen atom in aryl halides is in conjugation with π -electrons of the ring. This resonance imparts partial double bond character to C - X bond. Hence, the bond cleavage in haloarenes is difficult than haloalkanes since there is no double bond involved in them. Bond length: C - X bond in aryl halides is sp 2 -hybridised carbon atom with the greater s - character is more electronegative and hence the C - X bond is short. This shortening of bond length imparts stability to aryl halides and so bond cleavage becomes difficult.

How would you account for: Boiling points of aldehydes are lower than alcohols.

Intermolecular hydrogen bonding is present in alcohols, resulting in a strong bonding of a large number of molecules. As a result, the boiling points of alcohols are higher. The intermolecular force of attraction in aldehydes is dipole-dipole interaction, which is less than hydrogen bonding in alcohols. As a result, aldehydes have lower boiling points than alcohols.

Complete the reaction : CH 3 CH 2 Br → Dry ether N _____

Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing an even number of carbon atoms. Butane is formed when two moles of ethyl bromide react with two moles of sodium in the presence of dry or anhydrous ether. The following is a representation of the above reaction: 2 CH 3 CH 2 Br + 2 Na → Dry ether C 4 H 10 + 2 NaBr

Complete the following reaction: CH 3 CH 2 Br → AgCN _____

AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product. As silver cyanide is a covalent molecule, it prevents carbon from exhibiting nucleophilic activity because it is covalently bound to silver. As a result of the less nucleophilic nitrogen atom relative to carbon, the lone pair attacks bromoethane, the halogen that holds the carbon. The nitrogen carbon bond is formed as a result, and the product is ethyl carbylamine. The reaction is as follows: CH 3 CH 2 Br → AgCN CH 3 CH 2 NC + AgBr

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During the $β$ - elimination reaction of $2 -$ bromopentane in an alcoholic solution of $KOH$ results Pent $- 2 -$ ene as major product and Pent $- 1 -$ ene as minor product. State the rule to explain the reaction.

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