Let G be a finite group, and let B be a non-nilpotent block of G with respect to an algebraically closed field of characteristic 2. Suppose that B has an elementary abelian defect group of order 16 and only one simple module. The main result of this paper describes the algebra structure of the center of B. This is motivated by a similar analysis of a certain 3-block of defect 2 in [Kessar, 2012].Theorem 1.1. Let B be a non-nilpotent 2-block with elementary abelian defect group of order 16 and only one irreducible Brauer character. ThenIn particular, Z(B) has Loewy length 3.The paper is organized as follows. In the second section we consider the generalized decomposition matrix Q of B. Up to certain choices there are essentially three different possibilities for Q. A result by Puig [27] (cf. [9, Theorem 5.1]) describes the isomorphism type of Z(B) (regarded over O) in terms of Q. In this way we prove that there are at most two isomorphism types for Z(B). In the two subsequent sections we apply ring-theoretical arguments to the basic algebra of B in order to exclude one possibility for Z(B). Finally, we give some concluding remarks in the last section. Our notation is standard and can be found in [24,29].for any integer n ≥ 3. Now we will distinguish between the different cases that can occur for dim F (J 2 (A)/ J 3 (A)). We note that 2 ≤ dim F (J 2 (A)/ J 3 (A)) ≤ 4. The upper bound is clear by the preceding discussion, and if dim F (J 2 (A)/ J 3 (A)) = 1, then J 2 (A) ⊆ Z(A) by Lemma 3.3 which is a contradiction to dim F Z(A) = 8. The case dim F (J 2 (A)/ J 3 (A)) = 0 leads to J 2 (A) = 0 by Nakayama's Lemma and this is clearly false.proceed by distinguishing three subcases for an F -basis of J 2 (A)/ J 3 (A). More specifically there is always a basis of J 2 (A)/ J 3 (A) given by {x 2 + J 3 (A), d + J 3 (A)} for some d ∈ {xy, xz, yz}. This, however, implies J 3 (A) = F {x 3 , x 2 z} + J 4 (A) = F {x 2 z} + J 4 (A) and J 4 (A) = F {x 3 z} + J 5 (A) = J 5 (A).Hence, J 4 (A) = 0 by Nakayama's Lemma and therefore dim F A ≤ 1 + 3 + 2 + 1 = 7, a contradiction.(3): J 2 (A) = F {x 2 , yz} + J 3 (A) = F {x 2 , zy} + J 3 (A). We may assume that xy, xz ∈ F {x 2 } + J 3 (A) since otherwise we are in one of the previous two subcases. Using this we obtain J 3 (A) = F {x 3 , zxy, x 2 z, z 2 y} + J 4 (A) = F {x 3 , xyz, x 2 z} + J 4 (A) = F {x 3 } + J 4 (A). Hence, J 2 (A) ⊆ Z(A) by Lemma 3.3, and so dim F Z(A) ≥ dim F J 2 (A) = 12, a contradiction. We have thus shown that dim F (J 2 (A)/ J 3 (A)) = 2.Case (II.2): dim F (J 2 (A)/ J 3 (A)) = 3. Again, since x 2 / ∈ J 3 (A), there is always an F -basis of J 2 (A)/ J 3 (A) of the form {x 2 +J 3 (A), d 1 +J 3 (A), d 2 +J 3 (A)} for some d 1 , d 2 ∈ {xy, xz, yz}. Hence, we can proceed by distinguishing three subcases for a basis of J 2 (A)/ J 3 (A).