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Draw the products of the following reactions. If the products can exist as stereoisomers, show which stereoisomers are formed. a.

cis2pentene+HCl,cis-2-pentene + HCl,


trans2pentene+HCl,trans-2-pentene + HCl,


1ethylcyclohexene+H2O+H2SO4,1-ethylcyclohexene + H_2O + H_2SO_4,


2,3dimethyl3hexene+H2,Pd/C,2,3-dimethyl-3-hexene + H_2, Pd/C,


1,2dimethylcyclohexene+HCl,1,2-dimethylcyclohexene + HCl,


1,2dideuteriocyclohexene+H2,Pd/C,1,2-dideuteriocyclohexene + H_2, Pd/C,


3,3dimethyl1pentene+Br2/CH2Cl2,3,3-dimethyl-1-pentene + Br_2/CH_2Cl_2,


(E)3,4dimethyl3heptene+H2,Pd/C,(E)-3,4-dimethyl-3-heptene + H_2, Pd/C,


(Z)3,4dimethyl3heptene+H2,Pd/C,(Z)-3,4-dimethyl-3-heptene + H_2, Pd/C,


1chloro2ethylcyclohexene+H2,Pd/C1-chloro-2-ethylcyclohexene + H_2, Pd/C


Step 1
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Reaction of cis-2-pentene + HCl

In the first step, double bond of alkene act as nucleophile and attacks hydrogen attached to Cl and as a result carbocation forms. Point to be noted that hydrogen gets attached to carbon containing most hydrogen atoms so that positive charge gets stabilised. In this case two types of carbocation can form as shown in below figure as both carbocation is equally stable.

In the second step chlorine act as a nucleophile and attacks the positive charge. Now chlorine has two option to attack either from below the plane of carbocation or from above the carbocation. As a result two product will get formed corresponding to carbocation (above among two) as shown in figure. One product is R and other is S, therefore both are enantiomers. Also, since reactant doesn't have any stereocenter whereas reaction forms a product with an asymmetric center, therefore these are racemic mixture.

In other case where carbocation (below among two) is surrounded by bulky substituents, chlorine has only one option to attack and hence only one product forms in this case. Also as you can see that there is no chiral centre in product so no stereoisomers possible

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