Concept explainers
How would you prepare
each compound?
a.
b.
(a)
Interpretation:
The preparation of
Concept introduction:
Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. The reaction of alkyl halide with ammonia results in the formation of primary amine. Such reactions are commonly known as alkylation of ammonia.
Answer to Problem 25.54P
The preparation of
Explanation of Solution
The given compound is
The compound
(b)
Interpretation:
The preparation of
Concept introduction:
Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. Lithium aluminum hydride is a strong reducing agent. It is an inorganic compound which is used as a reducing agent in organic synthesis.
Answer to Problem 25.54P
The preparation of
Explanation of Solution
The given compound is
In the first step of reaction,
The conversion of
(c)
Interpretation:
The preparation of
Concept introduction:
Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. The conversion of nitroalkanes to primary amines occur by using the reducing agent
Answer to Problem 25.54P
The preparation of
Explanation of Solution
The given compound is
The conversion of
(d)
Interpretation:
The preparation of
Concept introduction:
Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. Lithium aluminum hydride is a strong reducing agent. It is an inorganic compound which is used as a reducing agent in organic synthesis.
Answer to Problem 25.54P
The preparation of
Explanation of Solution
The given compound is
The conversion of this to
(e)
Interpretation:
The preparation of
Concept introduction:
Synthesis is one of the major areas in the field of organic chemistry. It can be a simple one-step reaction or it may involve many steps. When an aldehyde or a ketone is converted to an amine, the process is known as reductive amination. The conversion takes place through the formation of an intermediate known as imine. The excess ammonia is used when a primary amine is synthesized by reductive amination.
Answer to Problem 25.54P
The preparation of
Explanation of Solution
The given compound is
The conversion of
Want to see more full solutions like this?
Chapter 25 Solutions
ALEKS 360 CHEMISTRY ACCESS
- What compounds are formed from the reaction of benzoyl chloride with the following reagents? a. sodium acetate b. water c. excess dimethylamine d. aqueous HCl e. aqueous NaOH f. cyclohexanol g. excess benzylamine h. 4-chlorophenol i. isopropyl alcohol j. excess aniline k. potassium formatearrow_forwardWhat is the reagent in this reaction? a. mCPBA, water, H+ b. H2, Pd c. NaBH4, water d. KMNO4arrow_forwardMatch the ff reagents: a. Br₂ in CH₂Cl₂ b.CrO₃ , H₂SO₄, acetone c. concentrated HCl with ZnCl₂ d. aqueous FeCl₃ e. aqueous NaHCO₃ f. ammoniacal AgNO₃ 1. 4-bromophenol from benzene 2. hexane from 1-hexene 3. 1-hexanol from hexanal 4. hexane from hexanoic acid 5. 2-hexanol from 2-methyl-2-hexanol 6. 1-hexene from 1-hexynearrow_forward
- what product is formed after recrystallizing a mixture of 4-phenylcyclohexanone and 4-phenylcyclohexanol?arrow_forward1. a. 4-methoxybenzoic acid is less or more polar than 4-methoxyacetophenone? explain why (WITHOUT DRAWINGS) b. 3′-chloro-4′-methoxyacetophenone is less or more polar than 4-methoxyacetophenone? explain why (WITHOUT DRAWINGS) 2. a. 4-methoxybenzoic acid has a higher melting point than 4-methoxyacetophenone. explain why? (WITHOUT DRAWINGS) b. 3′-chloro-4′-methoxyacetophenone has a higher melting point than 4-methoxyacetophenone. explain why? (WITHOUT DRAWINGS)arrow_forwardWhat is the best choice of reagents to achieve the following reaction? A. SOCl2 (1/2 equivalent) B. KOH (pellets) C. K2SO4 D. PhCO2Naarrow_forward
- What reagents would you use to convert methyl propanoate to the following compounds? a. isopropyl propanoate b. sodium propanoate c. N-ethylpropanamide d. propanoic acidarrow_forwardDraw a structural formula for the product formed by treating butanal with each reagent. (a) LiA1H4LiA1H4 followed by H2OH2O (b) NaBH4NaBH4 in CH3OH/H2O (c) H2/Pt (d) Ag(NH3)2+in NH3/H2O (e) H2CrO4, heat (f) HOCH2CH2OH,HClarrow_forwardDraw the products formed when p-methylaniline (p-CH3C6H4NH2) is treated with each reagent. a. HCl b. CH3COCl c. (CH3CO)2O d. excess CH3I e. (CH3)2C = O f. CH3COCl, AlCl3 g. CH3CO2H h. NaNO2, HCl i. Part (b), then CH3COCl, AlCl j. CH3CHO, NaBH3CNarrow_forward
- Draw the structure of each compound.(a) o-nitroanisole (b) 2,4-dimethoxyphenol (c) p-aminobenzoic acid(d) 4-nitroaniline (e) m-chlorotoluene (f) p-divinylbenzene(g) p-bromostyrene (h) 3,5-dimethoxybenzaldehyde (i) tropylium chloride(j) sodium cyclopentadienide (k) 2-phenylpropan-1-ol (l) benzyl methyl ether(m) p-toluenesulfonic acid (n) o-xylene (o) 3-benzylpyridinearrow_forwardCH3CH2CH2CH2Cl + 1) KCN 2) LiAlH4 3) H2O yields ___________. cyclopentylnitrile pentan-1-amine butan-1-nitrile pentan-1-nitrile 1-chloro-butan-1-nitrile Thank you!arrow_forwardGive the products formed when benzaldehyde and benzoic acid are treated with the given reagents. a. Tollen’s reagentb. phenylhydrazine, H+c. HCNd. NH2OHe. 1 mole H2, Nif. 1 mole CH3OH, H+g. LiAlH4 then H2O, H+h. 2 moles CH3OH, H+i. CH3MgCl, then H2O, H+j. H2Oarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning