a. in first step of the mechanism bromide ion liberated to form the secondary carbocation as shown below
b.
the geometry of the carbocation is trigonal planar with bond angle of 120 degree
c.
after first step hydrogen shift occur to form more stable tert-carbocation, later methanol attack to form the desired product.
d.
the formed product will be achiral compound, therefore, no stereoisomer is possible. since the methoxy contains carbon has two similar groups it is achiral. for chiral molecule we need carbon centre with four different group
The reaction show is an reaction Show the first step in this reaction CH,OH b What...
II. The reaction shown is an SN 1 reaction. a. Show the first step in this reaction. сн. + CH3OH Br b. What is the geometry at the reaction center in the intermediate step? C. Redraw the intermediate from step a, add the curved arrows and show the final step along with the resulting products. d. How many stereoisomers result from this reaction? Is the final product chiral? Why or why not? If the product is chiral, draw all stereoisomers...
organic chemistry
II. The reaction shown is an S1 reaction. a. Show the first step in this reattion CH Cн,он What is the geometry at the reaction center in the intermediate step? b. Redraw the intermediate from step a, add the curved arrows and show the final step along with the resulting products How many stereaisomers result from this reactionn? is the final product chiral? Why or why not? If the product is chiral draw all stereaisomers formed d.
a) Look at the reaction diagram for this reaction (Figure 9.2 in your textbook) and answer the following questions. 1) Is this diagram consistent with the rate law? Why or why not? 2) Experimental evidence shows that tertiary alkyl halides react the fastest and primary alkyl halides react the slowest with nucleophiles in this reaction. Explain why this is consistent with the reaction diagram in fig. 9.2. Stereochemistry of S 2 and S1 Reactions 1. The reaction shown is an...
First box is to add in curved
arrows that illustrate the first step of this mechanism. Second box
is Draw the two intermediates that form and show curved arrows
depicting the next step. Third is Draw the intermediate and the
small molecule with which it will react. Show curved arrows
depicting the next step. Fourth box is Draw the intermediate and
the small molecule with which it will react. Show curved arrows
depicting the next step.
05 Question (1 point)...
The Claisen condensation converts 2 molecules of an ester into a B-keto ester. The reaction starts with the ester in an alkoxide/alcohol solution and is worked up with acid to form the neutral B-keto ester product. :0 O: 1. -OCH2CH3/CH3CH20H 2 2. H30 OH Show the curved-arrow mechanism for the Claisen condensation of ethyl ethanoate treated with ethoxide ion. Include all formal charges and nonbonding electrons. In each step, draw only the species that react in that step. If an...
Show the mechanism for the following reaction conducted at –5 °C
in CCl4: cyclohexene bromine yields a dibromocyclohexane Draw
structures – including charges and electrons – and add curved
arrows. Details count.
Map do cyclohexene + bromine yields a dibromocyclohexane Draw structures - including charges and electrons - and add curved arrows. Details count. Add curved arrows to the first step. 1l Draw each species (organic and inorganic) resulting from the previous step. Include charges and nonbonding electrons. Add curved...
1.2.3.4.5.6.7.8.Show the curved arrow mechanism for the reaction between ethoxide and methanol to give ethanol and the methoxide ion. 1st attempt Jual See Periodic Table See Hint OH-Ö: Add the missing curved arrow notation.The carbon-metal bond in organometallic Grignard reagents exhibits significant covalent character. However, we can treat these compounds as electron-rich carbanions because of the large difference in electronegativity between carbon and magnesium. These reagents are great to form carbon-carbon bonds but must be kept in an anhydrous environment...
Map.com C,H,COCH, 1.NaBH, > ? 2.H,O+ Add curved arrows to show the mechanism of the first step of the reaction. Draw the charged organic intermediate product. Include nonbonding electrons and charges. Omit the counterion. click to edit Nath—B—H → continued below Draw the final product. * continued from above
2 problems
REVIEW Topics [References) CH HC OCH Br2 CH3 B H3C CH,OH H₂C CH₂ Electrophilic addition of hypohalous acids to alkenes yields a 1,2-haloalcohol called a halohydrin. Halohydrin formation, however, does not result from the addition of HOBr, for example. Instead the addition is done indirectly by reaction of the alkene with Br, in the presence of water. The reaction also works with Cl2 to give chlorohydrins instead of bromohydrins. The reaction proceeds through a cyclic intermediate known as...
Hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) yields 2-methylpropan-2-ol. (CH),CBr + 24,0 → (CH),COH + H20* + Br Give the Sn1 mechanism. Draw structures - including electrons and charges - and add curved arrows. Details count. Step 1 - Draw 2-bromo-2-methylpropane - Add one curved arrow. Draw the step 1 products: 1 organic species; 1 inorganic species. Map Step 2 . Reproduce the organic product of step 1. - Add one curved arrow to show water reacting with the organic species. Draw...