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大学有机化学英文版答案(2)

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导读: C H 3C H 2C H 2C H 2C lC H 3 C H 2 C H (M g B r)C H 3 H 2O C H 3C H 2C H 2C H 3 7.36 How might you prepare the following molecules using a nucleophilic substitution reaction at some step?(c ) CH3 C H

C H 3C H 2C H 2C H 2C lC H 3 C H 2 C H (M g B r)C H 3 H 2O C H 3C H 2C H 2C H 3

7.36 How might you prepare the following molecules using a nucleophilic substitution reaction at some step?(c ) CH3 C H 3O C C H 3 CH3

-O C C HCH3

CH3

3

+

C H 3I

大学有机化学英文版课后习题答案

7.48 The SN2 reaction can occur intramolecularly, meaning within the same molecule. What product would you expect from treatment of 4-bromo-1-butanol with base? Ob a se B rC H 2 C H 2 C H 2 C H 2 O H[ B rC H 2 C H 2 C H 2 C H 2 O

-

N a+ ]

7.53 Identify the following reactions as either SN1,SN2,E1,E2Br

(a)

CHCH3

C H=C H2 KOH

(b)

Br CHCH3 KOH

OCH3

CHCH3

E1

SN1

8.26Draw structural corresponding to the following IUPAC names(d ) tra n s -3 -e th ylc yc lo h e x a n o l(e ) 4 -a llyl-2 -m e th o x ylp h e n o lC H 3O HO C H 2C H = C H 2

HO

C H 2C H 3

8.27 Name the following compounds according to IUPAC rules:(b ) C H 3C H C H C H 2C H 3 HO C H 2C H 2C H 3(c ) Ph H OH H

3-ethyl-2-hexanol

cis-3-phenylcyclopentenol

大学有机化学英文版课后习题答案

8.33 Predict the product(s) of the following transformations:(a ) C H 2O H PCC CHO

8.43 Predict the product(s) of the following reactionsO (a ) HO OH N a 2C r 2O 7

8.59 Identify the reagents a through d in the following scheme:OH CH3 Na C H 3I OCH3 CH3 KOH ROH CH3 C H 3C O 3H CH3 O

O

-O HH 2O OH CH3 H OH

大学有机化学英文版课后习题答案

Naming Branched Alkanes找含C最多的主链若C数相同,找含支链最多的主链

编号,值最小 确定取代基及个数 写出命名 Prefix-parent-suffix 二 三 四 合并相同取代基 取代基中第一个字母顺序小的基团 先写出

用逗号隔开数字

‘—’隔开取代基 的数字和名称

大学有机化学英文版课后习题答案

IUPAC Rules for Alkene and Cycloalkene Nomenclature1. The -ene suffix (ending) indicates an alkene or cycloalkene. 2. The longest chain chosen for the root name must include both carbon atoms of the double bond. 3. The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond is in the center of th

e chain, the nearest substituent rule is used to determine the end where numbering starts. 4. The smaller of the two numbers designating the carbon atoms of the double bond is used as the double bond locator. If more than one double bond is present the compound is named as a diene, triene or equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator number. 5. In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule. 6. Substituent groups containing double bonds are: H2C=CH– Vinyl group H2C=CH–CH2– Allyl group

大学有机化学英文版课后习题答案

IUPAC Rules for Alkyne Nomenclature1. The -yne suffix (ending) indicates an alkyne or cycloalkyne. 2. The longest chain chosen for the root name must include both carbon atoms of the triple bond. 3. The root chain must be numbered from the end nearest a triple bond carbon atom. If the triple bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts. 4. The smaller of the two numbers designating the carbon atoms of the triple bond is used as the triple bond locator. 5. If several multiple bonds are present, each must be assigned a locator number. Double bonds precede triple bonds in the IUPAC name, but the chain is numbered from the end nearest a multiple bond, regardless of its nature. 6. Because the triple bond is linear, it can only be accommodated in rings larger than ten carbons. In simple cycloalkynes the triple bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule. 7. Substituent groups containing triple bonds are: HC≡C– Ethynyl group HC≡CH–CH2– Propargyl group

大学有机化学英文版课后习题答案

大学有机化学英文版课后习题答案

Sub-Rules for IUPAC NomenclatureThe root chain is that which contains the maximum number of multiple bonds. If more than one such chain is found, the longest is chosen as the root. If the chains have equal length the one with the most double bonds is chosen.

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