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
- The reaction of 3,4-dimethyl-3-hexanol (3,4-dimethylhexan-3-ol) with HBr generates compound A as the major product. Treatment of compound A with a strong base gives two isomers of compound B as the major product, along with one isomer of compound C and one isomer of compound D as minor products, all of which have one double-bond equivalent. Identify compounds A, B, C, and D and give their names. By what mechanism does the reaction of 3,4-dimethyl-3-hexanol with HBr occur? By what mechanism does the reaction of A with strong base to form B occur? Propose reaction conditions for an alternative, one-step method for converting 3,4-dimethyl-3-hexanol directly to compound B.arrow_forward(a) When cis-1-bromo-2-methylcyclohexane undergoes an E2 reaction, two products (cycloalkenes) are formed. What are these two cycloalkenes, and which would you expect to be the major product? Write conformational structures showing how each is formed. (b) When rans-1-bromo-2-methylcyclohexane reacts in an E2 reaction, only one cyclo- alkene is formed. What is this product? Write conformational structures showing why it is the only product.arrow_forwardRearrange 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_forward
- The reaction of (S)-2-bromopentane with potassium cyanide to yield 2-methylpentanenitrile (2-cyanopentane) occurs via a nucleophilic substitution pathway. The reaction is 100% stereospecific. Please explain in complete words what this observation tells you about the mechanism of the reaction.arrow_forward(a) Which diastereomer of oct-4-ene yields a mixture of two enantiomers, (4R,5R)- and (4S, 5S)-4,5-dibromooctane on reaction with Br2? (b) Which diastereomer of oct-4-ene yields a single meso compound, (4R, 5S)-4,5-dibromooctane?arrow_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
- Compounds 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_forwardReaction 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_forwardCompound A Br₂, H₂O Compound B (C8H15BrO) + enantiomer CH₂O O Compound C + enantiomer Draw the structure of Compound B (watch out for stereochemistry), and mechanisms for its formation from Compound A, and its conversion to Compound C.arrow_forward
- Identify 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_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_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_forward