Organic Chemistry Smartwork5 for Organic Chemistry 07 This is a Molecule Drawing question

Organic Chemistry Smartwork5 for Organic Chemistry 07 This is a Molecule Drawing question / It is WOF Question 16 (1 point): Draw the structure of (E)-1-chlorobut-1-ene. 5th attempt Feedback

The Correct Answer and Explanation is:

Correct structure (line formula)
Use the double-bond shown horizontally so you can place the ligands in their correct stereochemical positions:

      Cl           H
       \         /
        C1 == C2
        /         \
       H         CH2-CH3

— the chlorine on C1 and the ethyl group (-CH2-CH3) on C2 lie on opposite sides of the C1=C2 π-bond, satisfying the (E) (“entgegen”, German for “opposite”) configuration. A compact, machine-readable form is Cl/C=C/CC, the E-configured SMILES.

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Why this is the right drawing

1. Identify the parent skeleton
   But-1-ene is a four-carbon chain (but-) with a double bond starting at C-1 (-1-ene). Write the backbone as
   C1=C2-C3-C4.
2. Place the substituent
   The name carries the prefix 1-chloro-, meaning a chlorine atom replaces one of the hydrogens on C-1, the same carbon already involved in the double bond. Temporarily, therefore, C-1 is attached to Cl, H and C-2; C-2 is attached to H, C-1 and C-3.
3. Assign CIP priorities
   For alkene stereochemistry we compare the two substituents attached to each vinylic carbon.
   • At C-1 the higher atomic-number atom is Cl (Z = 17) versus H (Z = 1), so Cl gets priority 1, H priority 2.
   • At C-2 we compare the first atoms of each substituent: a carbon of an ethyl group (Z = 6) versus H (Z = 1). Thus CH2-CH3 is priority 1, H is priority 2.
4. Determine E or Z
   If the two priority-1 groups appear on the same side of the double bond, the isomer is Z (“zusammen”, together). If they lie on opposite sides, the isomer is E. The name specifies (E), so your sketch must place the Cl and the ethyl group across from one another. A convenient way is to draw the double bond horizontally, Cl above C-1 and CH2CH3 below C-2 (or vice-versa); the hydrogens automatically occupy the remaining, opposite positions.
5. Check valencies
   Each sp² carbon in the alkene must have four bonds in total (counting the double bond as two). The completed structure you just drew satisfies this, confirming chemical correctness.

Because only one stereogenic π-bond exists and its configuration is explicitly defined as E, the molecule you have drawn—Cl/C=C/CC—is the single, correct answer expected by Smartwork.