Interpretation:
The reason corresponding to the formation of products in each of the given reactions is to be explained.
Concept Introduction:
▸ The molecules which are non-superimposable or not identical with their mirror images are known as chiral molecules.
▸ A pair of two mirror images which are non-identical is known as enantiomers which are optically active.
▸ The objects or molecules which are superimposable with their mirror images are achiral objects or molecules and these objects have a centre of symmetry or plane of symmetry.
▸ The achiral compounds in which plane of symmetry is present internally and consists of chiral centres are known as meso compounds but they are optically inactive.
▸ The stereoisomers which are non-superimposable on each other and not mirror images of each other are known as diastereomers.
▸ Chiral molecules are capable of rotating plane polarized light
▸ The molecules which are superimposable or identical with their mirror images are known as achiral molecules, and achiral molecules are not capable of rotating the plane-polarised light.
▸ Priority is given to all the four group attached to the chirality center.
▸ Priority is assigned on the basis of the
▸ If priority cannot be assigned according to
▸ After assigning priority to the four groups, rotate the molecule such that fourth priority group is away from the observer.
▸ Now, move from a to b to c; if the direction is clockwise, then the chiral center designated as
▸
▸ Z-isomers have the high priority group on the same side of the double bond whereas E-isomers have the high priority group on the opposite side of the double bond.
▸ In
▸ In alkenes, if the higher priority group on both the carbon is on the opposite side, configuration is termed as E-configuration.
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Organic Chemistry
- Rearrange the compounds of each of the following sets in increasing order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) l-Bromo-3-methylbutane, 2-Bromo-2-methyl butane, 3-Bromo-2- Methylbutane. (iii) 1 -Bromobutane, l-Bromo-2,2 -dimethylpropane, 1 -Bromo-2 -methyl butanearrow_forwardCompounds X and Y are both C7H15Cl products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C7H₁4 alkene. Both X and Y undergo an SN2 reaction with sodium iodide in acetone solution to give C7H15l products; in this reaction Y reacts faster than X. What is the structure of X? • Do not use stereobonds in your answer. • In cases where there is more than one possible structure for each molecule, just give one for each. . Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with + signs from the drop-down menu. наarrow_forwardGive the SN2 product(s) for the reaction of (R)-2-bromopentane with an excess of sodium methoxide. (R)-2-methoxypentane (S)-2-methoxypentane (S)-2-ethoxypentane (R)-2-ethoxypentane (R)-2-pentanol O(S)-2-pentanolarrow_forward
- Reaction of this bicycloalkene with bromine in carbon tetrachloride gives a trans dibro- mide. In both (a) and (b), the bromine atoms are trans to each other. However, only one of these products is formed. CH3 CH3 CH3 Br Br CH,Cl, + Br2 or Br Br (a) (b) Which trans dibromide is formed? How do you account for the fact that it is formed to the exclusion of the other trans dibromide?arrow_forwardDiphenylacetylene can be synthesized by the double dehydrohalogenation of 1,2-dibromo-1,2-diphenylethene. The sequence starting from (E)-1,2-diphenylethene consists of bromination to give the dibromide, followed by dehydrohalogenation to give a vinylic bromide, then a second dehydrohalogenation to give diphenylacetylene.(a) What is the structure, including stereochemistry, of the vinylic bromide?(b) If the sequence starts with (Z)-1,2-dibromo-1,2-diphenylethene, what is (are) the structure(s) of the intermediate dibromide(s)? What is the structure of the vinylic bromide?arrow_forwardIdentify the compound in each of the following pairs that reacts more rapidly in SN2 reactions: (a) 1-bromopentane or 3-bromopentane (b) 2-chloropentane or 2-fluoropentane (c) 2-bromopropane or 1-bromohexane (d) 1-chlorohexane or cyclohexyl chloridearrow_forward
- A chemist allows some pure (2S,3R)-3-bromo-2,3-diphenylpentane to react with a solution of sodium ethoxide(NaOCH2 CH3) in ethanol. The products are two alkenes: A (cis-trans mixture) and B, a single pure isomer. Under the same conditions, the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane gives two alkenes, A (cis-trans mixture) and C. Upon catalytic hydrogenation, all three of these alkenes (A, B, and C) give 2,3-diphenylpentane. Determine the structures of A, B, and C; give equations for their formation; and explain the stereospecificity of these reactions.arrow_forwardTreatment of propadiene (an allene) with hydrogen bromide produces 2-bromopropene as the major product. This suggests that the more stable carbocation intermediate is produced by the addition of a proton to Br HBr. H2C=C=CH, H3C CH2 a terminal carbon rather than to the central carbon. Propadiene 2-Bromopropene (a) Draw both carbocation intermediates that can be produced by the addition of a proton to the allene. (b) Explain the relative stabilities of those intermediates. Hint: Draw the orbital picture of the intermediates and consider whether the CH, groups in propadiene are in the same plane.arrow_forward3-bromo-1-pentene and 1-bromo-2-pentene undergo SN1 reaction at almost the same rate, but one is a secondary halide while the other is a primary halide. Explain your answer. Then, decide which the following compounds will react faster in an E2 reaction; trans-1-bromo-2-isopropylcyclohexane or cis-1-bromo-2-isopropylcyclohexane. and Explain your answer.arrow_forward
- When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1- butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base? base Br A. Less than 45% B. C. 45% Between 45% and 70% D. More than 70%arrow_forwardCompounds X and Y are both C,H15CI products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C-H14 alkene. Both X and Y undergo an SN2 reaction with sodium iodide in acetone solution to give C-H151 products; in this reaction Y reacts faster than X. What is the structure of X? • Do not use stereobonds in your answer. • In cases where there is more than one possible structure for each molecule, just give one for each. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with + signs from the drop-down menu. In (F ChemDoodlearrow_forwardPredict the structure (stereochemistry) of the dibromide products of a synaddition of bromine to trans-cinnamic acid. Predict the structure (stereochemistry) of the dibromide products of an antiaddition of bromine to trans-cinnamic acid.Label each product as threo or erytho.It may be helpful to construct molecular models of threo-and erytho-2,3-dibromo-3-phenylpropanioc acid.arrow_forward