
Concept explainers
(a)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
Molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image. But not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are superimposable.
Explanation of Solution
Molecules in the given pair are
In the above pair, both molecules have carbon as the central atom with one fluorine atom and three hydrogen atoms as an attachment. In the left molecule, fluorine is pointing towards the observer and one hydrogen atom is pointing away from the observer while in the second molecule fluorine is pointing away from the observer and one hydrogen atom is pointing towards the observer.
To check the molecule in different orientation the molecule present at right is rotated.
The molecule after the rotation has the exact similar orientation of the atoms as the left molecule in the given pair i.e. all atoms are lined up perfectly.
Thus both the molecules in the given pair are superimposable.
All atoms of both molecules are lined perfectly hence superimposable.
(b)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
Molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image, but not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are superimposable.
Explanation of Solution
Molecules in the given pair are
In given pair, both molecules are mirror images of each other. If the molecule at right is rotated, it gives the molecule exactly similar to the left one.
Thus both the molecules in the given pair are superimposable.
All atoms of both the molecules are lined perfectly hence superimposable.
(c)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
Molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image, but not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are nonsuperimposable.
Explanation of Solution
Molecules in the given pair are:
Two molecules given in the pair are the cis-trans isomers of the same molecule. To get the trans isomer from cis, rotation is needed around the double bond which is restricted. Hence the atoms in both molecules are not lined up perfectly, thus the given two molecules are nonsuperimposable.
All atoms of both molecules are not lined perfectly hence nonsuperimposable.
(d)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
The molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image, but not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are superimposable.
Explanation of Solution
Molecules in the given pair are:
Two molecules given in pairs are the mirror images of each other. If one of the two given molecules is rotated around the central carbon atom it gives the exact similar molecule as another.
Rotation of the molecule gives the molecule with a similar orientation as another molecule, thus two molecules are superimposable.
All atoms of both molecules are lined perfectly, hence superimposable.
(e)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
The molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image, but not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are superimposable.
Explanation of Solution
Molecules in the given pair are:
Both molecules in the given pair are isomers of the same molecules thus have the same attachments of the molecule with different orientations. Two chlorine atoms in the left molecule are pointing towards the observer while two chlorine atoms in the right molecule are pointing away from the observer. If one of the molecule rotates with
The molecule after the rotation of the left molecule has the exact similar orientation as the molecule at right, all atoms of both molecules are lined perfectly representing the molecules in the given pair are superimposable.
All atoms of both molecules are lined perfectly hence superimposable.
(f)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
Molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image, but not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are nonsuperimposable.
Explanation of Solution
Molecules in the given pair are
In the molecule at the left, both chlorine atoms are pointing towards the observer while in the molecule at right one chlorine atom is pointing towards the observer and one is pointing away from the observer. To get a similar orientation, rotation of the atom with the ring is needed which is restricted. Thus there is no orientation of both molecules lined perfectly, representing the molecules in the given pair are nonsuperimposable.
All atoms of both molecules are not lined perfectly, hence nonsuperimposable.
(g)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
Molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image. But not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are superimposable.
Explanation of Solution
Molecules in the given pair are
In the given pair both molecules are mirror images of each other having one chlorine atom pointing towards the observer and another pointing away from the observer.
The complete rotation of
The molecule after the rotation of the left molecule has the exact similar orientation as the molecule at right, all atoms of both molecules are lined perfectly representing the molecules in the given pair are superimposable.
All atoms of both molecules are lined perfectly hence superimposable.
(h)
Interpretation:
Whether the molecules in the given pair are superimposable or nonsuperimposable is to be determined.
Concept introduction:
The molecules are nonsuperimposable if there is no orientation in which all atoms of both molecules can be lined up perfectly (i.e., superimposed). Every molecule can have a mirror image. But not every molecule can have a nonsuperimposable mirror image.

Answer to Problem 5.1P
The given two molecules are nonsuperimposable.
Explanation of Solution
Molecules in the given pair are
In the molecule at left both chlorine atoms are pointing towards the observer while in the molecule at right one chlorine atom is pointing towards the observer and one is pointing away from the observer. To get a similar orientation, rotation of the atom with the ring is needed which is restricted. Thus no orientation of both molecules lined perfectly, representing the molecules in the given pair are nonsuperimposable.
All atoms of both molecules are not lined perfectly hence nonsuperimposable.
Want to see more full solutions like this?
Chapter 5 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Draw the major product of this reaction. Ignore inorganic byproducts. ○ O 1. H₂O, pyridine 2. neutralizing work-up a N W X 人 Parrow_forward✓ Check the box under each molecule that has a total of five ẞ hydrogens. If none of the molecules fit this description, check the box underneath the table. tab OH CI 0 Br xx Br None of these molecules have a total of five ẞ hydrogens. esc Explanation Check caps lock shift 1 fn control 02 F2 W Q A N #3 S 80 F3 E $ t 01 205 % 5 F5 & 7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility FT * 8 R T Y U כ F6 9 FIG F11 F D G H J K L C X V B < N M H option command P H + F12 commandarrow_forwardDraw the major product of this reaction. Ignore inorganic byproducts and the carboxylic acid side product. O 1. CHзMgBr (excess) 2. H₂O ✓ W X 人arrow_forward
- If cyclopentyl acetaldehyde reacts with NaOH, state the product (formula).arrow_forwardDraw the major product of this reaction. Ignore inorganic byproducts. N S S HgCl2, H2SO4 く 8 W X Parrow_forwardtab esc く Drawing the After running various experiments, you determine that the mechanism for the following reaction occurs in a step-wise fashion. Br + OH + Using this information, draw the correct mechanism in the space below. 1 Explanation Check F2 F1 @2 Q W A os lock control option T S # 3 80 F3 Br $ 4 0105 % OH2 + Br Add/Remove step X C F5 F6 6 R E T Y 29 & 7 F D G H Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Ce A F7 DII F8 C Ո 8 * 9 4 F10 F C J K L C V Z X B N M H command P ge Coarrow_forward
- Indicate compound A that must react with ethylbenzene to obtain 4-ethylbenzene-1-sulfonic acid. 3-bromo-4-ethylbenzene-1-sulfonic acid.arrow_forwardPart 1 of 2 Draw the structure of A, the minor E1 product of the reaction. esc I Skip Part Check H₂O, D 2 A + Click and drag to start drawing a structure. -0- F1 F2 1 2 # 3 Q A 80 F3 W E S D F4 $ 4 % 5 F5 ㅇ F6 R T Y F G X 5 & 7 + Save 2025 McGraw Hill LLC. All Rights Reserved. DII F7 F8 H * C 80 J Z X C V B N 4 F9 6arrow_forwardFile Preview The following is a total synthesis of the pheromone of the western pine beetle. Such syntheses are interesting both because of the organic chemistry, and because of the possibility of using species specific insecticides, rather than broad band insecticides. Provide the reagents for each step. There is some chemistry from our most recent chapter in this synthesis, but other steps are review from earlier chapters. (8 points) COOEt COOEt A C COOEt COOEt COOH B OH OTS CN D E See the last homework set F for assistance on this one. H+, H₂O G OH OH The last step is just nucleophilic addition reactions, taking the ketone to an acetal, intramolecularly. But it is hard to visualize the three dimensional shape as it occurs. Frontalin, pheromone of the western pine beetlearrow_forward
- For the reaction below: 1. Draw all reasonable elimination products to the right of the arrow. 2. In the box below the reaction, redraw any product you expect to be a major product. C Major Product: Check + ◎ + X ง © Cl I F2 80 F3 I σ F4 I F5 NaOH Click and drawing F6 A 2025 McGraw Hill LLC. All Rights E F7 F8 $ # % & 2 3 4 5 6 7 8 Q W E R T Y U A S D F G H Jarrow_forwardCan I please get help with this graph. If you can show exactly where it needs to pass through.arrow_forwardN Draw the major product of this reaction. Ignore inorganic byproducts. D 1. H₂O, pyridine 2. neutralizing work-up V P W X DE CO e C Larrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
